$\begin{aligned} & \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x-1\right)}{5-(4+\cos x)}\{\sqrt{5}+\sqrt{4+\cos x}\} \\ & =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x+1\right)}{1-\cos x} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\} \\ & =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x-1\right)}{2 \sin ^2\left(\frac{x}{2}\right)} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\} \\ & =\lim _{x \rightarrow 0} \frac{\left(\frac{9^x-1}{x}\right)\left(\frac{3^x-1}{x}\right)}{2 \times \frac{\sin ^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{1}{4}} \cdot\{\sqrt{5}+\sqrt{4+\cos x}\}\end{aligned}$
$\begin{aligned} & =\frac{\log 9 \cdot \log 3}{\frac{1}{2}}\{\sqrt{5}+\sqrt{5}\} \\ & =2 \log 9 \cdot \log 3 \cdot(2 \sqrt{5}) \\ & =8 \sqrt{5}(\log 3)^2\end{aligned}$